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# If (t- 8) is a factor of t^2 - kt - 48, then k=

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If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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24 Jan 2014, 02:53
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If (t- 8) is a factor of t^2 - kt - 48, then k=

(A) - 6
(B) - 2
(C) 2
(D) 6
(E) 14

Problem Solving
Question: 57
Category: Algebra Second-degree equations
Page: 69
Difficulty: 600

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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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24 Jan 2014, 02:53
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SOLUTION

If (t- 8) is a factor of t^2 - kt - 48, then k=

(A) - 6
(B) - 2
(C) 2
(D) 6
(E) 14

$$(t - 8)$$ is a factor of $$t^2 - kt - 48$$ means that $$t = 8$$ is a solution of the equation $$t^2 - kt - 48 = 0$$.

Substitute$$t = 8$$ to get the value of k: $$8^2 - 8k - 48 = 0$$ --> $$k = 2$$.

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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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24 Jan 2014, 04:55
Factoring the expression:
(t - 8) (t - ?), since (t - 8) is a factor, the other bracket has to be (t + 6).
The net result is -2. So (B)? Not sure of the answer because I got lost with the signs a bit.
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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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24 Jan 2014, 22:27
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Put the value of t-8=0, or t=8 in equation t^2+kt+48=0: Solving we get k=2.
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If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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24 Jan 2014, 23:46
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If (t- 8) is a factor of $$t^2 - kt - 48$$, then k=

(A) - 6
(8) - 2
(C) 2
(0) 6
(E) 14

It is given that $$(t - 8)$$ is a factor of the quadratic expression $$t^2 - kt - 48$$
Hence, we need to find the other factor of $$-48$$ such that the sum of factors is$$-(\frac{-k}{1})=k$$.

Thus, $$(8) * x = -48$$

Or, $$x = -6$$

So, $$k = 8 + (- 6) = 2$$

Last edited by arunspanda on 01 Nov 2014, 02:02, edited 1 time in total.
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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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25 Jan 2014, 11:39
SOLUTION

If (t- 8) is a factor of t^2 - kt - 48, then k=

(A) - 6
(B) - 2
(C) 2
(D) 6
(E) 14

$$(t - 8)$$ is a factor of $$t^2 - kt - 48$$ means that $$t = 8$$ is a solution of the equation $$t^2 - kt - 48 = 0$$.

Substitute$$t = 8$$ to get the value of k: $$8^2 - 8k - 48 = 0$$ --> $$k = 2$$.

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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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25 Feb 2014, 18:44
-8 x ?? = -48
?? = 6, so the other factor is 6
(t-8)(t+6) would be the two factors
-k = -8+6
-k = -2
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If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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22 May 2014, 23:38
Vieta's formulas applied to quadratic: x1+x2= -b/a & x1*x2=c/a
From what is given in the question: -8+x2=-k and -8*x2=-48
so, x2=6 and thus -8+6=-k i.e. k=2
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Last edited by NoHalfMeasures on 11 Nov 2015, 20:45, edited 1 time in total.
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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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30 Jun 2015, 15:40
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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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30 Jun 2015, 21:27
since (t-8) is a factor, then: (8)^2 - 8k - 48 = 0,
-8k = -16
k = 2.
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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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15 May 2016, 10:04
In this question, I didn't understand this step:

(t−8)(t−8) is a factor of t2−kt−48t2−kt−48 means that t=8t=8 is a solution of the equation t2−kt−48=0t2−kt−48=0.

Why do we assume both equations are equal to 0?
What is the rationale of this step? It would be great if someone could explain with examples, or with videos.
Re: If (t- 8) is a factor of t^2 - kt - 48, then k=   [#permalink] 15 May 2016, 10:04
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